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For every sigma-algebra G and every (suitably integrable) random variables X and Y such that Y is G-measurable, one has E(XY)=E(E(X|G)Y). This is one of the two conditions that characterize E(X|G), the other one being that E(X|G) is G-measurable. In particular, for every event A, E(Y;A)=E(P(A|G)Y).

Apply this to G your sigma-algebra $\mathcal{F}_{=t}$, A the event $A_1$ and Y the conditional probability of $A_2$ conditionally on G.

Thank you! In definition of E(X|G), I remember it is E(X)=E(E(X|G); A) for any A in G. So is E(XY)=E(E(X|G)Y) for any G-measurable random variable Y really part of its definition? Why is E(XY)=E(E(X|G)Y) true?
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EthanMay 4 '11 at 11:08

It is equivalent to ask that E(X;A)=E(E(X|G);A) for every A in G or to ask that E(XY)=E(E(X|G)Y) for every (bounded) G-measurable random variable Y. The latter implies logically the former but it is a standard exercise to prove that the two are in fact equivalent.
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DidMay 4 '11 at 11:42

From the former to the latter, is it proved by using simple functions to approximate Y when Y is nonnegative?
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EthanMay 4 '11 at 11:56